To determine the probability of getting 1 or fewer successes in a binomial experiment with \( N = 16 \) trials and a success probability of \( p = 0.2 \), we can use the binomial probability formula. The formula for the probability of getting exactly \( k \) successes is:
\[ P(X = k) = \binom{N}{k} p^k (1-p)^{N-k} \]
Where:
- \( \binom{N}{k} \) is the binomial coefficient, which represents the number of ways to choose \( k \) successes from \( N \) trials.
- \( p \) is the probability of success on each trial.
- \( (1-p) \) is the probability of failure on each trial.
We want to find \( P(X \leq 1) = P(X = 0) + P(X = 1) \).
Step 1: Calculate \( P(X = 0) \)
\[ P(X = 0) = \binom{16}{0} (0.2)^0 (0.8)^{16} \] \[ = 1 \cdot 1 \cdot (0.8)^{16} \] \[ = (0.8)^{16} \] \[ \approx 0.0407 \quad (\text{using a calculator for } (0.8)^{16}) \]
Step 2: Calculate \( P(X = 1) \)
\[ P(X = 1) = \binom{16}{1} (0.2)^1 (0.8)^{15} \] \[ = 16 \cdot (0.2) \cdot (0.8)^{15} \] \[ = 16 \cdot 0.2 \cdot (0.8)^{15} \] \[ = 3.2 \cdot (0.8)^{15} \] \[ \approx 3.2 \cdot 0.0512 \quad (\text{using a calculator for } (0.8)^{15}) \] \[ \approx 0.1638 \]
Step 3: Calculate \( P(X \leq 1) \)
Now we add \( P(X = 0) \) and \( P(X = 1) \): \[ P(X \leq 1) = P(X = 0) + P(X = 1) \] \[ \approx 0.0407 + 0.1638 = 0.2045 \]
Final Answer
Thus, the probability \( P(1 \text{ or fewer successes}) \) is approximately: \[ \boxed{0.2045} \]